A224868 -id:A224868 - cp's OEIS frontend

A227728 a(1) = greatest k such that H(k) - H(2) < 1/1 + 1/2; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(2); and for n>2, a(n) = greatest k such that H(k) - H(a(n-1)) < H(a(n-1)) - H(a(n-2)), where H = harmonic number.

10, 43, 179, 740, 3054, 12599, 51971, 214376, 884278, 3647546, 15045706, 62061794, 255997704, 1055960840, 4355715996, 17966823308, 74111062350, 305699536774, 1260975134078, 5201376179830, 21455073484758, 88499689759294, 365050956038686, 1505792854949114

Offset: 1

Clark Kimberling, Jul 29 2013

Suppose that x and y are positive integers and that x <= y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A224868, (x,y) = (1,2) and a(n) = c(n+1) for n >= 1. It appears that the sequence a(n) is linearly recurrent with signature (5,-4,2,-2,2,-2,2,-2,2,-2). See A227729 and A225815 for limits of the sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1).

			The first three values (a(1),a(2),a(3)) = (10,43,179) match the beginning of the following inequality chain (and partition of the harmonic numbers): 1/1 + 1/2 > 1/3 + 1/4 + ... + 1/10 > 1/11 + ... + 1/43 > 1/44 + ... + 1/179 > ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A224820, A224868, A225815, A227729.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A227728 *)
N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 50]  (* A227729 *)
N[Table[a[n]/a[n - 1], {n, 2, z, 25}], 50]  (* A225815 *)

a(n) = 5*a(n-1) - 4*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + 2*a(n-7) - 2*a(n-8) + 2*a(n-9) - 2*a(n-10) (conjectured).

Empirical g.f.: -x*(4*x^9-4*x^8+3*x^7-4*x^6+3*x^5-4*x^4+3*x^3-4*x^2+7*x-10) / ((x-1)*(2*x^9+2*x^7+2*x^5+2*x^3+4*x-1)). - Colin Barker, Mar 23 2015

Definition corrected by Clark Kimberling, Jan 14 2017

A227965 a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Original entry on oeis.org

11, 53, 249, 1164, 5435, 25371, 118428, 552798, 2580343, 12044484, 56221045, 262427666, 1224955522, 5717827134, 26689578960, 124581175389, 581517950673, 2714399875409, 12670230858892, 59141894115145, 276061555506087, 1288595564424512, 6014885070144844

Offset: 1

Clark Kimberling, Aug 01 2013

Suppose that x and y are positive integers and that x <=y. Let a(1) = least k such that H(y) - H(x-1) < H(k) - H(y); let a(2) = least k such that H(a(1)) - H(y) < H(k) - H(a(1)); and for n > 2, let a(n) = least k such that greatest such H(a(n-1)) - H(a(n-2)) < H(k) - H(a(n-1)). The increasing sequences H(a(n)) - H(a(n-1)) and a(n)/a(n-1) converge. For what choices of (x,y) is the sequence a(n) linearly recurrent?

For A227965, (x,y) = (1,2); H(a(n)) - H(a(n-1)) approaches a limit 1.540684... given by A227966, and a(n)/a(n-1) approaches a limit 4.6677834... given by A227967. It is unknown whether the sequence a(n) is linearly recurrent.

			The first two values (a(1),a(2)) = (11,53) match the beginning of the following inequality chain (and partition of the harmonic numbers):  1/1 + 1/2 < 1/3 + ... + 1/11 < 1/12 + ... + 1/53 < ...

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A224820, A224868, A227728, A227966, A227967.

z = 300; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 1; y = 2;
a[1] = Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}];
m = Map[a, Range[z]] (* A227965 *)
t = N[Table[h[a[t]] - h[a[t - 1]], {t, 2, z, 25}], 60]
Last[RealDigits[t, 10]]  (* A227966 *)
t = N[Table[a[t]/a[t - 1], {t, 2, z, 50}], 60]
Last[RealDigits[t, 10]]  (* A227967 *)
(* A227965,  Peter J. C. Moses, Jul 12 2013*)

A227816 a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Original entry on oeis.org

16, 41, 103, 257, 640, 1592, 3958, 9839, 24457, 60792, 151107, 375596, 933591, 2320556, 5768028, 14337143, 35636731, 88579473, 220175161, 547272407, 1360312788, 3381224518, 8404448844, 20890289891, 51925381404, 129066913288, 320811665802, 797416799492

Offset: 1

Clark Kimberling, Jul 31 2013

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. It appears that for many choices of (x,y), the sequence c(n) is linearly recurrent with signature of length less than 10; (3,6) is not one of them. H(c(n)) - H(c(n-1)) and c(n)/c(n-1) approach limits given by A227817 and A227818.

			The first two values (a(1),a(2)) = (16,41) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 3 ):
1/3 + 1/4 + 1/5 + 1/6 > 1/7 + ... + 1/16 < 1/17 + ... + 1/41 <  ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).

Cf. A224820, A224868, A227728, A227817, A227818.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 6; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* (A227804) Peter J. C. Moses, Jul 23 2013 *)

A227804 a(1) = greatest k such that H(k) - H(8) < H(8) - H(4); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(8), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Original entry on oeis.org

15, 27, 48, 85, 150, 264, 464, 815, 1431, 2512, 4409, 7738, 13580, 23832, 41823, 73395, 128800, 226029, 396654, 696080, 1221536, 2143647, 3761839, 6601568, 11584945, 20330162, 35676948, 62608680, 109870575, 192809419, 338356944, 593775045, 1042002566

Offset: 1

Clark Kimberling, Jul 31 2013

Suppose that x and y are positive integers and that x <=y. Let c(1) = y and c(2) = greatest k such that H(k) - H(y) < H(y) - H(x); for n > 2, let c(n) = greatest such that H(k) - H(c(n-1)) < H(c(n-1)) - H(c(n-2)). Then 1/x + ... + 1/c(1) > 1/(c(1)+1) + ... + 1/(c(2)) > 1/(c(2)+1) + ... + 1/(c(3)) > ... The decreasing sequences H(c(n)) - H(c(n-1)) and c(n)/c(n-1) converge. For what choices of (x,y) is the sequence c(n) linearly recurrent?

For A227804, (x,y) = (5,8); it appears that the sequence a(n) is linearly recurrent with signature (3,-3,2,-1), that H(c(n)) - H(c(n-1)) approaches a limit 0.56239..., and that c(n)/c(n-1) approaches the constant 1.75487... given at A109134.

			The first three values (a(1),a(2),a(3)) = (10,43,179) match the beginning of the following inequality chain (and partition of the harmonic numbers H(n) for n >= 5 ):  1/5 + 1/6 + 1/7 + 1/8 > 1/9 + ... + 1/15 < 1/16 + ... + 1/27 < 1/28 + ... + 1/48 > ...

Clark Kimberling, Table of n, a(n) for n = 1..1000

Cf. A001008, A002805 (numerator and denominator of harmonic numbers).

Cf. A224820, A224868, A227728.

z = 100; h[n_] := h[n] = HarmonicNumber[N[n, 500]]; x = 3; y = 5; a[1] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[y] - h[x - 1], {w, 1}, WorkingPrecision -> 400]]; a[2] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[1]] - h[y], {w, a[1]}, WorkingPrecision -> 400]]; Do[s = 0; a[t] = -1 + Ceiling[w /. FindRoot[h[w] == 2 h[a[t - 1]] - h[a[t - 2]], {w, a[t - 1]}, WorkingPrecision -> 400]], {t, 3, z}]; m = Map[a, Range[z]] (* A227804, Peter J. C. Moses, Jul 23 2013 *)

a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - a(n-4) (conjectured).

G.f.: (15 - 18 x + 12 x^2 - 8 x^3)/(1 - 3 x + 3 x^2 - 2 x^3 + x^4) (conjectured).

cp's OEIS Frontend

A227728 a(1) = greatest k such that H(k) - H(2) < 1/1 + 1/2; a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(2); and for n>2, a(n) = greatest k such that H(k) - H(a(n-1)) < H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A227965 a(1) = least k such that 1 + 1/2 < H(k) - H(2); a(2) = least k such that H(a(1)) - 1/2 < H(k) -H(a(1)), and for n > 2, a(n) = least k such that H(a(n-1)) - H(a(n-2)) > H(k) - H(a(n-1)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A227816 a(1) = greatest k such that H(k) - H(6) < H(6) - H(3); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(6), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A227804 a(1) = greatest k such that H(k) - H(8) < H(8) - H(4); a(2) = greatest k such that H(k) - H(a(1)) < H(a(1)) - H(8), and for n > 2, a(n) = greatest k such that H(k) - H(a(n-1)) > H(a(n-1)) - H(a(n-2)), where H = harmonic number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula